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#1 2007-02-11 01:13:01

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Try this

This is a little problem I thought up while I was bored, I can't seem to solve it, but you have a go:

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#2 2007-02-11 01:41:06

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Try this

I think I've got it. I've got an answer that sounds feasible anyway.
I probably used an inefficient and long-winded way of working it out, so I don't want to put up all of my lengthy workings, but here's my answer in case anyone else wants to check:


Why did the vector cross the road?
It wanted to be normal.

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#3 2007-02-11 02:00:21

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Try this

(looking at pic)

Last edited by luca-deltodesco (2007-02-11 02:12:28)


The Beginning Of All Things To End.
The End Of All Things To Come.

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#4 2007-02-11 02:17:01

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Try this

Huh. That's pretty much the same method I used, and yet it somehow took up 3 pages of scribbles. Ah well. roll


Why did the vector cross the road?
It wanted to be normal.

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#5 2007-02-11 23:03:33

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: Try this

Could you please explain how you got x and y? I don't really understand tongue. Thanks.

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#6 2007-02-12 01:43:19

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Try this

x is a sector of the circle. You get its area by finding the angle to see what fraction of the circle it is, and then taking that fraction of the circle's area.

luca has worked out angle b by using cos θ = adj/hyp = 5/6, and the angle of the sector is double angle b. To find the area as a fraction of the circle's area, you divide the angle by 2π.

So then we have 2b/2π = b/π. Multiplying this by the circle's area will get the area of the sector. b/π * π*6² = b*6² = 36b.

y is a triangle. luca has used Pythagoras to find that half of the triangle's base is √11, and then multiplied by 2 to get the total base. Because the square has side length 10, it's fairly easy to see that the triangle's height is 5.

Therefore, it's area is (2√11*5)/2 = 5√11.

And then z is the triangle subtracted from the sector, and the shaded regions are the quadruple of z.


Why did the vector cross the road?
It wanted to be normal.

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